Relation between strong convergence in $L^{p}$ and weak convergence in $H_{0}^{1}(\Omega)$

Question

Let $\{u_{n}\}_{n\in\mathbb{N}}$ be a bounded sequence in $H_{0}^{1}(\Omega)$ for a bounded interval $\Omega \subset \mathbb{R}$. By weak compactness of Hilbert Space, we can extract a subsequence of $\{u^{1}_{n_{k}}\}_{k\in\mathbb{N}}$ such that $u^{1}_{n_{k}}\to u^{1}$ weakly in $H_{0}^{1}(\Omega)$.

On the other hand, by Rellich Theorem, we can obtain another subsequence $\{u^{2}_{n_{k}} \}_{k\in\mathbb{N}}$ such that $u^{2}_{n_{k}}\to u^{2}$ strongly in $L^{p}(\Omega)$ for $2<p<\infty$.

My question is how can we show that $u^{1} = u^{2}$? At the very least, I want to obtain a subsequence which satisfies both properties.

My attempt so far is first to extract subsequence $\{u^{1}_{n_{k}} \}_{k\in\mathbb{N}}$ from $\{u_{n}\}_{n\in\mathbb{N}}$. Then, I extract subsequence $\{u^{2}_{n_{k}} \}_{k\in\mathbb{N}}$ as another subsequence from $\{u^{1}_{n_{k}} \}_{k\in\mathbb{N}}$. Is this enough to ensure that $u^{1} = u^{2}$? If it is enough, in what sense can I say that $u^{1} = u^{2}$?

Any help is much appreciated! Thank you very much

Answer 1

Recall the following result.

Let $X,Y$ be Hilbert spaces and $T:X \to Y$ a compact operator.

Let $\{x_n\}$ be a sequence in $X$ such that $x_n \to x$ weakly in $X$.

Then, it holds that $Tx_n \to Tx$ in $Y$.

Proof.

Let $y_n=Tx_n$. It is enough to show that every subsequence of $y_n$ has a convergent subsequence which converges to $Tx$.

Let $y_{n_k}$ be a subsequence of $y_n$. Since $T$ is a compact operator, $y_{n_k}$ is a sequence in a compact metric space. Therefore, there exists a convergent subsequence $y_{n_{k_l}}$. We denote by $y \in Y$ the limit.

It remains to show $y=Tx$. Since $x_{n_{k_l}} \to x$ weakly in $X$, $y_{n_{k_l}}=Tx_{n_{k_l}} \to Tx$ weakly in $Y$. Hence $y=Tx$.